Question: Find the zeros of the function. Enter the solutions from least to greatest. $f(x) = (x - 2)^2 - 9$ $\text{lesser }x = $
$\begin{aligned} (x - 2)^2 - 9&= 0 \\\\ (x-2)^2&=9 \\\\ \sqrt{(x-2)^2}&=\sqrt{9} \end{aligned}$ $\begin{aligned} x-2&=\pm3 \\\\ x&=\pm3+2 \\ \phantom{(x - 2)^2 - 9}& \\ x=-1&\text{ or }x=5 \end{aligned}$ In conclusion, $\begin{aligned} \text{lesser }x &= -1 \\\\ \text{greater } x &= 5 \end{aligned}$